A Chemist wanted to determine the percentage of calcium carbonate in a sea shell. He collected a number of sea shells which he crushed into a fine powder.
To 2.65 grams of sea shell powder he added 25.0mL of 5.0M Hydrochloric acid.
When all the fizzing ceased, he filtered the remaining mixture into a conical flask and carefully washed the residue.
Finally he titrated the filtrate with 2.20M Sodium Hydroxide and 46.3mL of base was required.
Calculate the percentage of calcium carbonate in the sea shells.
The idea here is that you add an excess of HCl to the sea shells and that completely reacts with ALL of the calcium carbonate. The excess HCl is titrated with NaOH.
CaCO3 + 2HCl -> CaCl2 + 2H2O
mols HCl added initially = M x L = 0.025 x 5 = 0.125.
How much HCl was in excess? That's 0.0436 x 2.2 = 0.0959
Here's what we have.
0.125 mols HCl initially
-0.0959 mols xs HCl
------------
0.0291 mols HCl used by the CaCO3.
Convert mols HCl to mols CaCO3.
0.0291mols HCl x (1 mol CaCO3/2 mols HCl) = 0.0291/2 = ?mols CaCO3
Convert to grams. g = mols CaCO3 x molar mass CaCO3.
Then %CaCO3 = (mass CaCO3/mass sample)*100 - ?
To calculate the percentage of calcium carbonate in the sea shells, we need to analyze the reaction that took place between the hydrochloric acid and the sea shell powder.
The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is as follows:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to form 1 mole of calcium chloride, 1 mole of carbon dioxide, and 1 mole of water.
First, let's calculate the number of moles of hydrochloric acid used:
Moles of HCl = (Volume of HCl used in liters) x (Molarity of HCl)
Given:
Volume of HCl used = 25.0 mL = 0.0250 L
Molarity of HCl = 5.0 M
Moles of HCl = (0.0250 L) x (5.0 mol/L) = 0.125 mol
Since the reaction ratio between calcium carbonate and hydrochloric acid is 1:2, we can deduce that the number of moles of calcium carbonate is half the number of moles of hydrochloric acid used.
Moles of CaCO3 = 0.125 mol / 2 = 0.0625 mol
Next, let's calculate the number of moles of sodium hydroxide used in the titration reaction.
Moles of NaOH = (Volume of NaOH used in liters) x (Molarity of NaOH)
Given:
Volume of NaOH used = 46.3 mL = 0.0463 L
Molarity of NaOH = 2.20 M
Moles of NaOH = (0.0463 L) x (2.20 mol/L) = 0.102 mol
Since the balanced chemical equation shows that 1 mole of sodium hydroxide reacts with 1 mole of calcium carbonate, the number of moles of calcium carbonate present in the shell powder is equal to the moles of sodium hydroxide used in the titration.
Moles of CaCO3 = 0.102 mol
Now, let's calculate the molar mass of calcium carbonate:
Molar mass of CaCO3 = (Molar mass of Ca) + (Molar mass of C) + (3 x Molar mass of O)
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol
Finally, we can calculate the mass of calcium carbonate in the sea shell powder:
Mass of CaCO3 = Moles of CaCO3 x Molar mass of CaCO3
Mass of CaCO3 = 0.102 mol x 100.09 g/mol = 10.2098 g
Now, we can calculate the percentage of calcium carbonate in the sea shells:
Percentage of CaCO3 = (Mass of CaCO3 / Mass of sea shell powder) x 100
Given:
Mass of sea shell powder = 2.65 g
Percentage of CaCO3 = (10.2098 g / 2.65 g) x 100 = 385.83%
Therefore, the percentage of calcium carbonate in the sea shells is approximately 385.83%.
To calculate the percentage of calcium carbonate in the sea shells, we need to determine the amount of calcium carbonate that reacted with the hydrochloric acid and was titrated with sodium hydroxide.
1. Calculate the number of moles of hydrochloric acid used:
- 25.0 mL of 5.0 M hydrochloric acid is used.
- Convert the volume to liters: 25.0 mL = 0.025 L.
- Calculate the number of moles: moles = concentration (M) x volume (L).
- Moles = 5.0 M x 0.025 L = 0.125 moles.
2. Determine the balanced chemical equation for the reaction between hydrochloric acid and calcium carbonate:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the equation, we see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid.
3. Calculate the number of moles of calcium carbonate that reacted:
- From the balanced equation, 2 moles of HCl react with 1 mole of CaCO3.
- Therefore, the moles of CaCO3 = 0.125 moles / 2 = 0.0625 moles.
4. Calculate the molar mass of calcium carbonate:
- The molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol.
5. Calculate the mass of calcium carbonate that reacted:
- Mass = moles x molar mass = 0.0625 moles x 100.09 g/mol = 6.25625 g.
6. Calculate the percentage of calcium carbonate in the sea shells:
- The original mass of the sea shell powder used was 2.65 grams.
- Percentage = (mass of CaCO3 / mass of sea shell powder) x 100%
- Percentage = (6.25625 g / 2.65 g) x 100% = 235.9%.
Therefore, the percentage of calcium carbonate in the sea shells is approximately 235.9%.